# Class 9 RD Sharma Solutions – Chapter 18 Surface Area and Volume of a Cuboid and Cube – Exercise 18.2 | Set 2

**Question 14. The dimensions of a cinema hall are 100 m, 50 m, 18 m. How many persons can sit in the hall, if each person requires 150 m**^{3} of air?

^{3}of air?

**Solution:**

Length of hall = 100 m

Breadth of hall = 50 m

Height of hall = 18 m

So, the Volume of the hall = l Ã— b Ã— h

= 100 Ã— 50 Ã— 18 = 90000 m

^{3}Also, the volume of air required per person = 150 m

^{3}So we can say the number of persons who can sit in the hall will be given by, Volume of Hall/Volume of air required by one person

= 90000/150 = 600 persons

Hence, 600 personscan sit in the hall, if each person requires 150 m^{3}of air

**Question 15. Given that 1 cubic cm of marble weighs 0.25 kg, the weight of the marble block 28 cm in width, and 5 cm thick is 112 kg. Find the length of the block.**

**Solution:**

Given,

Breadth of marble block = 28 cm

Height of marble block = 5 cm

Let

lbe the length of the blockSo, the Volume of the block = l Ã— b Ã— h

= l Ã— 28 Ã— 5 = 140l cm

^{3}Since weight of 1cm

^{3}is 0.25 kgSo, weight of marble block will be 0.25 Ã— 140l kg

The total weight of marble is 112 kg

So, we can say

112 = 0.25 Ã— 140l

l = 3.2 cm

Hence, the length of marble block is 3.2 cm

**Question 16. A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of a fluid can be placed in it? Also, find the volume of the wood used in it.**

**Solution:**

Given,

External length of box = 25 cm

External breadth of box = 18 cm

External height of box = 15 cm

So, the volume of the external box = l Ã— b Ã— h

= 25 Ã— 18 Ã— 15 = 6750 cm

^{3}Now, the internal dimensions of the box will be given as,

The length will be 25 -(2 Ã— 2) = 21 cm

The breadth will be 18 -(2 Ã— 2) = 14 cm

The height will be 15 -(2 Ã— 2) = 11 cm

Now, the internal volume of the cuboid is given as l Ã— b Ã— h

= 21 Ã— 14 Ã— 11 = 3234 cm

^{3}So, we can say the volume of liquid that can be filled in the cuboid box = 3234 cm

^{3}And the volume of wood needed or used = External volume of the box -Internal volume of the box

= 6750 -3234 = 3516 cm

^{3}

Hence, the volume of liquid that can be filled inside the box is 3234 cm^{3}and the volume of wood needed or used is 3516 cm^{3}

**Question 17. The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm Ã— 3 cm Ã— 0.75 cm can be put in this box?**

**Solution:**

Given,

External length of wooden box = 48 cm

External breadth of wooden box = 36 cm

External height of wooden box = 30 cm

Also, the thickness of wood is 1.5 cm

So, the internal dimensions of the box can be given as,

The length will be 48 -(2 Ã— 1.5) = 45 cm

The breadth will be 36 -(2 Ã— 1.5) = 33 cm

The height will be 30 -(2 Ã— 1.5) = 27 cm

Now, the internal volume of the cuboid is given as l Ã— b Ã— h

= 45 Ã— 33 Ã— 27 = 40095 cm

^{3}The dimensions of brick are given as 6 cm Ã— 3 cm Ã— 0.75 cm

So, the volume of brick will be l Ã— b Ã— h

= 6 Ã— 3 Ã— 0.75 = 13.5 cm

^{3}Therefore, we can say the number of bricks that can be put inside the wooden box = Internal volume of the wooden box/Volume of one brick

= 40095/13.5 = 2970 bricks

Hence, the number of bricks that can be put inside the wooden box is 2970.

**Question 18. How many cubic centimeters of iron are there in an open box whose external dimensions are 36 cm, 25 cm**,** and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cubic cm of iron weighs 15 gms. Find the weight of the empty box in kg.**

**Solution:**

The external length of iron = 36 cm

The external breadth of iron = 25 cm

The external height of iron = 16.5 cm

So, the external volume is given as = l Ã— b Ã— h

= 36 Ã— 25 Ã— 16.5 = 14850 cm

^{3}Also, the internal dimensions of the iron can be given as,

The length will be 36 -(2 Ã— 1.5) = 33 cm

The breadth will be 25 -(2 Ã— 1.5) = 22 cm

The height will be 16.5 -1.5 = 15 cm

Now, the internal volume of the cuboid is given as l Ã— b Ã— h

= 33 Ã— 22 Ã— 15 = 10890 cm

^{3}Therefore, the weight of iron = External volume -Internal volume

= 14850 -10890 = 3960 cm

^{3}And the weight of iron is given as, 15 Ã— 3960 = 59400 gm = 59.4 kg

Hence, the weight of iron is 59.4 kg

**Question 19. A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel.**

**Solution:**

Given, the edge of cube = 9 cm

So, we can say the volume of a cube = (edge)

^{3}= 9

^{3}= 729 cm^{3}The dimensions of the base of the rectangular vessel are 15 cm Ã— 12 cm

So, the area of rectangular base = l Ã— b

= 15 Ã— 12 = 180 cm

^{2}Now, the rise in water level in the vessel will be given as,

Volume of cube/Area of base of rectangular vessel = 729/180 = 4.05 cm

Hence, the rise of water level in the vessel is about 4.05 cm

**Question 20. A rectangular container, whose base is a square of side 5cm, stands on a horizontal table and holds water up to 1cm from the top. When a solid cube is placed in the water it is completely submerged, the water rises to the top**,** and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.**

**Solution:**

Let

abe the edge of each cubeFrom the Questiontion, we can conclude that the volume of the cube will be equal to the sum of the volume of water present inside the tank and the volume of water that overflowed from the tank

So, the volume of cube = The volume of water present inside the tank + The volume of water that overflowed from the tank

a

^{3}= (5 Ã— 5 Ã— 1) + 2a

^{3}= 27a = 3 cm

So, the volume of the cube is 27 cm

^{3}And edge will be 3 cm

Hence, the volume and edge of the cube are 27 cm^{3}and 3 cm respectively.

**Question 21. A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7m deep and the earth taken out is spread evenly on the field. By how many meters is the level of the field raised? Give the answer to the second place of decimal.**

**Solution:**

The dimensions of earth dug out = 50 m Ã— 40 m Ã— 7 m

So, the volume of earth dug out = l Ã— b Ã— h

= 50 Ã— 40 Ã— 7 = 14000 m

^{3}Let the height of the field rise by

hmAlso, we know that volume of the field will be equal to the volume of earth dug out

So, 200 Ã— 150 Ã— h = 14000

h = 0.47 m

Hence, thelevel of the field raised by 0.47 m

**Question 22. A field is in the form of a rectangular length of** **18m and width of 15m. A pit 7.5m long, 6m broad**,** and 0.8m deep, is dug in a corner of the field and the earth taken out is spread over the remaining area of the field. Find out the extent to which the level of the field has been raised.**

**Solution:**

Let the height of the field rise by

hmThe volume of earth taken out of the pit will be, l Ã— b Ã— h

= 7.5 Ã— 6 Ã— 0.8 = 36 m

^{3}Also, the earth is spread out on the field whose area can be given by 18 Ã— 15 -7.5 Ã— 6

= 225 m

^{2}As we also know, the volume of earth taken out from pit = Area of the field Ã— h

So, 36 = 225 Ã— h

h = 16 cm

Hence, thelevel of the field has been raised by 16 cm

**Question 23. A rectangular tank is 80 m long and 25 m broad. Water flows into it through a pipe whose cross-section is 25 cm ^{2}, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes?**

**Solution:**

Let

hcm be the rise in water levelSo, the volume of the water tank is given as l Ã— b Ã— h

= 8000 Ã— 2500 Ã— h cm

^{3}Also, the cross-sectional area of the pipe is 25 cm

^{2}From the Questiontion, we can say the water coming out of the pipe forms a cuboid of base area 25 cm

^{2}, and length will be equal to the distance travelled with 16 kph in 45 minutes.So, length = 16000 Ã— 100 Ã— 45/60 cm

So, we conclude the volume of water coming out of the pipe in 45 minutes = 25 Ã— 16000 Ã— 100 Ã— 45/60

Also, the volume of water in the tank will be equal to the volume of water coming out of the pipe

Therefore, 8000 Ã— 2500 Ã— h = 25 Ã— 16000 Ã— 100 Ã— 45/60

h = 1.5 cm

Hence, the level of water rises by 1.5 cm in the tank in 45 minutes.

**Question 24. Water in a rectangular reservoir having a base 80 m by 60 m is 6.5m deep. **At** what time can the water be pumped by a pipe of which the cross-section is a square of side 20 cm if the water runs through the pipe at the rate of 15km/hr.**

**Solution:**

Given, the flow of water in the pipe is 15km/hr

= 15000 m/hr

The volume of water coming out of the pipe in an hour can be given by,

20/100 Ã— 20/100 Ã— 15000 = 600 m

^{3}Now the volume of the tank = l Ã— b Ã— h

= 80 Ã— 60 Ã— 6.5 = 31200 m

^{3}Therefore, the time taken to fill the empty tank = Volume of tank/Volume of coming out of the pipe in an hour

= 31200/600 = 52 hours

Hence, thewater must be pumped for 52 hours

**Question 25. A village having a population of 4000 requires 150 liters of water per head per day. It has a tank measuring 20 m Ã— 15 m Ã— 6 m. For how many days will the water of this tank last?**

**Solution:**

Given, the length of the tank = 20 m

The length of tank = 15 m

The length of tank = 6 m

So, the volume/capacity of tank = l Ã— b Ã— h

= 20 Ã— 15 Ã— 6 = 1800 m

^{3}= 1800000 litres

The amount of water consumed by people in the village = 150 Ã— 4000 litres = 600000 litres

Let the water in the tank last for

adaysSo, the amount of water consumed by all people of the village in

adays = Volume of tankSo, a Ã— 600000 = 1800000

a = 3 days

Hence, the water will last for 3 days

**Question 26. A child playing with building blocks, which are of the shape of the cubes, has built a structure. If the edge of each cube is 3cm, find the volume of the structure built by the child.**

**Solution:**

Volume of cube = edge

^{3}= 3

^{3}= 27 cm^{3}Total number of the cube in the figure = 15

So, the volume of figure = 15 Ã— 27 = 405 cm

^{3}

**Question 27. A godown measures 40 m Ã— 25 m Ã— 10 m. Find the maximum number of wooden crates each measuring 1.5 m Ã— 1.25 m Ã— 0.5 m that can be stored in the godown.**

**Solution:**

Given, the dimensions of the godown are,

40 m Ã— 25 m Ã— 10 m

So, the volume of godown = l Ã— b Ã— h

= 40 Ã— 25 Ã— 10 = 10000 m

^{3}Also, the dimensions of the wooden crate are,

1.5 m Ã— 1.25 m Ã— 0.5 m

So, the volume of the wooden crate = l Ã— b Ã— h

= 1.5 Ã— 1.25 Ã— 05 = 0.9375 m

^{3}The number of wooden crates that can be stored = Volume of godown/Volume of one wooden crate

= 10000/0.9375 = 10666.66 wooden crates

Hence, approx 10667 wooden crates can be stored in the godown.

**Question 28. A wall of length 10 m was to be built across open ground. The height of the wall is 4 m and the thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm Ã— 12 cm Ã— 8 cm, how many bricks would be required?**

**Solution:**

Given, the dimensions of the wall are,

1000 cm Ã— 24 cm Ã— 400 cm

So, the volume of the wall = l Ã— b Ã— h

= 1000 Ã— 24 Ã— 400 = 9600000 cm

^{3}Also, the dimensions of the brick are,

24 cm Ã— 12 cm Ã— 8 cm

So, the volume of the one brick = l Ã— b Ã— h

= 24 Ã— 12 Ã— 8 = 2304 cm

^{3}The number of bricks required = Volume of wall/Volume of one brick

= 9600000/2304 = 4166.66 bricks

Hence, approx 4167 bricks are required to build the given wall.